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# Clearing Denominators in Rational Equations

## Multiply both sides by a common multiple

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Practice Clearing Denominators in Rational Equations

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Workers' Comp
Teacher Contributed

## Real World Applications – Algebra I

### Topic

How can we lower workers’ compensation in companies using Algebra?

### Student Exploration

Most companies have a Human Resources department that handles all of the logistics about the company’s workers. One of their biggest responsibilities is working with employees when they get hurt on the job. When this happens, the company offers workers’ compensation. For one particular company, the human resources advisor created algebraic relationships to represent the number of people that were hurt at one of the branches of the company. He concluded that the function that represents this relationship is I0(x)=500x\begin{align*}I_0(x)= \frac{500}{x}\end{align*}. The advisor also created a plan to implement after the initial training in attempts to reduce the number of injuries at work. With additional extensive training and check-ups with employees, he concluded that this model would produce I1(x)=500(x+1)\begin{align*}I_1(x)= \frac{500}{(x + 1)}\end{align*} less injuries. We can represent the difference between the two models to see if the number of injuries would be effective or not.

We can represent this revised model by taking the difference between the two relationships as one rational expression.

I0I1=500x500x+1\begin{align*}I_0-I_1= \frac{500}{x} - \frac{500}{x+1}\end{align*}

Since we’re subtracting two rational expressions, we have to make sure we have a common denominator, and then subtract. In this case, our common denominator is x(x+1)\begin{align*}x(x + 1)\end{align*}, so we do the following:

In the picture above, we found the common denominator by making the denominators the same, and then we simplified the numerator.

What does this resulting expression mean?

We can input some values into a table for this function and make graph the relationship. It would look like this:

For this graph, the x\begin{align*}x\end{align*} values represent the number of hours of training, and the y\begin{align*}y\end{align*} values represent the number of injuries per year. This graph clearly shows that that number of injuries has been drastically reduced over time. What if we were to compare the two graphs together separately? What would that graph tell us? See below.

The graphs are really close together, so it’s a little difficult to compare. So, let’s take a look at the tables that represent the two relationships.

# of hours of training (x)\begin{align*}(x)\end{align*} Without plan With plan
5 100 83.33333333
10 50 45.45454545
15 33.33333333 31.25
20 25 23.80952381
25 20 19.23076923
30 16.66666667 16.12903226
50 10 9.803921569
100 5 4.95049505
200 2.5 2.487562189

Looking at this table, we can see that the plan will definitely make a difference. For every number of hours of training, we can see that the rightmost column has significantly lower numbers. The plan will work!

We can also use the formula I0I1=500x500x+1\begin{align*}I_0-I_1= \frac{500}{x}- \frac{500}{x+1}\end{align*} or 500x2+x\begin{align*}\frac{500}{x^2+x}\end{align*} to make predictions of the model, assuming that it’s been proven effective for the past and present.

Let’s use 500x2+x\begin{align*}\frac{500}{x^2+x}\end{align*} and find out when this would equal 5 incidents over the course of a year. We can find out how many trainings would be needed so that we shouldn’t have no more than 5 injuries. We set our expression 500x2+x\begin{align*}\frac{500}{x^2+x}\end{align*} equal to 5 and then solve for x\begin{align*}x\end{align*}.

When we set this equal to 5, we can look at solving this two different ways. We can look at this in the sense like we’re solving a proportion and 5 is the same as 51\begin{align*}\frac{5}{1}\end{align*}.

500x2+x=51\begin{align*}\frac{500}{x^2+x}=\frac{5}{1}\end{align*}

Or, we can solve it like we’re rationalizing the equation and trying to clear the denominator in this rational equation. We would take 500x2+x=5\begin{align*}\frac{500}{x^2+x}=5\end{align*} and multiply both sides of the equation by x2+x\begin{align*}x^2+x\end{align*}. Essentially, either of these two methods will yield the same next step.

Let’s work with the proportion 500x2+x=51\begin{align*}\frac{500}{x^2+x}=\frac{5}{1}\end{align*} and multiply both sides of the equation by 1 and x2+x\begin{align*}x^2+x\end{align*}. We now have 500=5x2+5x\begin{align*}500=5x^2+5x\end{align*}. It’s a quadratic! Since we’re trying to find our x\begin{align*}x-\end{align*}values, we can solve this using the quadratic formula after setting the equation equal to zero. Check all of the steps below:

After using the quadratic formula, we’ve found two answers. Would both answers make sense in this situation? And what do these answers mean?

Since we’re trying to find out how many trainings it would take to ensure that the number of incidents are no more than 5, 9.5 would make sense. Offering -10.5 trainings wouldn’t make sense!

### Extension Investigation

Research different companies that would represent a similar type of model. Why would this company have this type of model? Why did you choose this company? Would this model be limited to injuries? What about production of certain goods?

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